Given that x,y `in` R, solve (2 + 3i) `x^2` - (3-2i)y = 2x - 3y + 5i

Given that x,yinR solve x^2-y^2-i(2x+y)=4i

Given that x, y in R . Solve: x/(1+2i)+y/(3+2i)=(5+6i)/(8i-1)

Given that x, y in R . Solve: x/(1+2i)+y/(3+2i)=(5+6i)/(8i-1)

Given that x, y in R . Solve: x/(1+2i)+y/(3+2i)=(5+6i)/(8i-1)

Given that x, y in R . Solve (x)/(1+2i) + (y)/(3+2i)= (5+6i)/(8i-1)

Given that x,y in R. Solve: (x)/(1+2i)+(y)/(3+2i)=(5+6i)/(8i-1)

If 3x + y i= x + 2y , then 2x - y =

Solve the following equations, x, y in R : (x + iy) / (2 + 3i) = 7 - i

x,y in R, solve :4x+3xy+(2xy-3x)i=4y-((x^(2))/(2))+(3xy-2y)i

Find the values of the real numbers x and y, if the complex numbers (3 - i) x - (2-i) y + 2i + 5 and 2x + (-1 + 2i)y + 3 + 2i are equal